Inca o limita... (Own)
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Virgil Nicula
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Inca o limita... (Own)
Sa se arate ca \( \lim_{n\to\infty}n^{\alpha_n}\ , \) unde \( \alpha_n=\frac {\ln n!}{nH^2_n} \) si \( H_n=\frac 11+\frac 12+\ldots +\frac 1n\ . \)
Last edited by Virgil Nicula on Fri Mar 14, 2008 11:43 pm, edited 3 times in total.
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Cezar Lupu
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Vom folosi binecunoscuta inegalitate, care se demonstreaza usor prin inductie:
\( \left(\frac{n}{e}\right)^{n}< n! <\left(\frac{n}{e}\right)\cdot (n+1), \forall n\geq 5 \). Logaritmand, vom obtine ca
\( n(ln n-1)< ln n! < n(ln n-1)+ln(n+1) \)
Astfel, vom avea ca
\( \frac{n(ln n-1)}{H_{n}^{2}}<\frac{ln n!}{H_{n}^{2}}< \frac{n(ln n-1)}{H_{n}^{2}} +\frac{ln(n+1)}{H_{n}^{2} \)
Notam cu \( a_{n}=\frac{ln n!}{H_{n}^{2}} \). Mai departe, avem
\( (\sqrt[n]{n})^{a_{n}}=e^{ln n^{\frac{a_{n}}{n}}=e^{\frac{a_{n} ln n}{n} \).
Prin urmare, avem ca \( \lim_{n\to\infty}(\sqrt[n]{n})^{a_{n}}=e^{\lim_{n\to\infty}{\frac{a_{n} ln n}{n} \).
Acum fosim acea dubla inegalitate pentru a prinde intr-un cleste sirul nostru.
Intr-adevar, avem
\( \frac{ln n(ln n-1)}{H_{n}^{2}}<\frac{a_{n}ln n}{n}<\frac{ln n(ln n-1)}{H_{n}^{2}}+\frac{ln n\cdot ln(n+1)}{nH_{n}^{2}} \).
Acum, este destul de clar, ca limita sirului de mai sus este \( 1 \)din moment ce \( \lim_{n\to\infty} H_{n}=\infty \) si \( \lim_{n\to\infty}\frac{H_{n}}{ln n}=1 \). Pentru ultima limita se aplica Stolz-Cesaro sau se tine cont ca \( \lim_{n\to\infty}{H_{n}}-{ln n}=\gamma \)
In cocluzie, limita sirului \( (\sqrt[n]{n})^{a_{n}}=e \).
\( \left(\frac{n}{e}\right)^{n}< n! <\left(\frac{n}{e}\right)\cdot (n+1), \forall n\geq 5 \). Logaritmand, vom obtine ca
\( n(ln n-1)< ln n! < n(ln n-1)+ln(n+1) \)
Astfel, vom avea ca
\( \frac{n(ln n-1)}{H_{n}^{2}}<\frac{ln n!}{H_{n}^{2}}< \frac{n(ln n-1)}{H_{n}^{2}} +\frac{ln(n+1)}{H_{n}^{2} \)
Notam cu \( a_{n}=\frac{ln n!}{H_{n}^{2}} \). Mai departe, avem
\( (\sqrt[n]{n})^{a_{n}}=e^{ln n^{\frac{a_{n}}{n}}=e^{\frac{a_{n} ln n}{n} \).
Prin urmare, avem ca \( \lim_{n\to\infty}(\sqrt[n]{n})^{a_{n}}=e^{\lim_{n\to\infty}{\frac{a_{n} ln n}{n} \).
Acum fosim acea dubla inegalitate pentru a prinde intr-un cleste sirul nostru.
Intr-adevar, avem
\( \frac{ln n(ln n-1)}{H_{n}^{2}}<\frac{a_{n}ln n}{n}<\frac{ln n(ln n-1)}{H_{n}^{2}}+\frac{ln n\cdot ln(n+1)}{nH_{n}^{2}} \).
Acum, este destul de clar, ca limita sirului de mai sus este \( 1 \)din moment ce \( \lim_{n\to\infty} H_{n}=\infty \) si \( \lim_{n\to\infty}\frac{H_{n}}{ln n}=1 \). Pentru ultima limita se aplica Stolz-Cesaro sau se tine cont ca \( \lim_{n\to\infty}{H_{n}}-{ln n}=\gamma \)
In cocluzie, limita sirului \( (\sqrt[n]{n})^{a_{n}}=e \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.